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burhan101
 3 years ago
Trig identity.. I don't understand this one at all
burhan101
 3 years ago
Trig identity.. I don't understand this one at all

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burhan101
 3 years ago
Best ResponseYou've already chosen the best response.0how do i make the left side into one big fraction

burhan101
 3 years ago
Best ResponseYou've already chosen the best response.0\[\huge \frac{ 1 }{ sinx+1 }  \frac{ 1 }{ sinx1 } = \frac{ 2 }{ 1\sin^2x}\]

burhan101
 3 years ago
Best ResponseYou've already chosen the best response.0typed it wrong wooops

dpaInc
 3 years ago
Best ResponseYou've already chosen the best response.2get common denominators on the left side: \(\huge \frac{ 1 }{ sinx+1 }\cdot (\color {red}{\frac{ sinx1 }{ sinx1 }})  \frac{ 1 }{ sinx1 }\cdot( \color {red}{\frac{ sinx+1 }{ sinx+1 }}) = \frac{ 2 }{ 1\sin^2x} \)

dpaInc
 3 years ago
Best ResponseYou've already chosen the best response.2so the left side in simplified form will be: \(\huge \frac{(sinx1)(sinx+1)}{sin^2x1} \) can you simplify this?

burhan101
 3 years ago
Best ResponseYou've already chosen the best response.0no that's where i get stuck how would i subtract the numerator

dpaInc
 3 years ago
Best ResponseYou've already chosen the best response.2just use distributive property to get rid of the parenthesis...

burhan101
 3 years ago
Best ResponseYou've already chosen the best response.0@dpaInc i get this \[\huge \frac{ \sin^2x+sinxsinx1 }{ sinx^21 }\]

dpaInc
 3 years ago
Best ResponseYou've already chosen the best response.2how did you get that? all you're doing is subtracting....: \(\huge \frac{(sinx1)(sinx+1)}{sin^2x1} \) =\(\huge \frac{1\cdot (sinx1)1\cdot (sinx+1)}{sin^2x1} \) =\(\huge \frac{sinx1sinx1}{sin^2x1} \) =\(\huge \frac{2}{sin^2x1} \) =\(\huge \frac{2}{1sin^2x} \)

burhan101
 3 years ago
Best ResponseYou've already chosen the best response.0sinx1sinx1 = 2 is that a trig rule or what ? :S
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